A block of mass m = 1 kg slides with velocity v = 6 m/s on a frictionless horizontal surface and collides with a uniform vertical rod and sticks to it as shown. The rod is pivoted about O and swings as a result of the collision making angle $$\theta $$ before momentarily coming to rest. If the rod has mass M = 2 kg, and length $$l$$ = 1 m, the value of $$\theta $$ is approximately :
(take g = 10 m/s²)

A particle of mass m with an initial velocity $$u\widehat i$$ collides perfectly elastically with a mass 3 m at rest. It moves with a velocity $$v\widehat j$$ after collision, then, v is given by :

A particle of mass m is projected with a speed u from the ground at an angle $$\theta = {\pi \over 3}$$ w.r.t. horizontal (x-axis). When it has reached its maximum height, it collides completely inelastically with another particle of the same mass and velocity $$u\widehat i$$ . The horizontal distance covered by the combined mass before reaching the ground is:

A rod of length L has non-uniform linear mass
density given by $$\rho $$(x) = $$a + b{\left( {{x \over L}} \right)^2}$$ , where a
and b are constants and 0 $$ \le $$ x $$ \le $$ L. The value
of x for the centre of mass of the rod is at :